- #1
johnaphun
- 14
- 0
I've been asked to prove that the following distribution is a member of the generalised exponential family of distributions.
f(y;β) = (ky2β(y+k))/((β+3)(y+2k)(y+1)1/2)
I know that i have to transform the equation into the form
f(y) = exp{(yθ-bθ)/a∅ +c(y,∅)}
and that to do this i should take the exponential of the log
exp{log(f(y;β))}
I understand how to do this for the more simple distributions (poisson, binomial etc) however i always struggle with more complicated ones. Are there any tips or anything to look out for when answering this type of question.
So far i have
exp[(y+k)log(βy)-(y+2k)log(β+3)+log ky - (1/2)log(y+1)]
but I'm pretty sure that's not correct
f(y;β) = (ky2β(y+k))/((β+3)(y+2k)(y+1)1/2)
I know that i have to transform the equation into the form
f(y) = exp{(yθ-bθ)/a∅ +c(y,∅)}
and that to do this i should take the exponential of the log
exp{log(f(y;β))}
I understand how to do this for the more simple distributions (poisson, binomial etc) however i always struggle with more complicated ones. Are there any tips or anything to look out for when answering this type of question.
So far i have
exp[(y+k)log(βy)-(y+2k)log(β+3)+log ky - (1/2)log(y+1)]
but I'm pretty sure that's not correct